Question

The number of six-digit natural numbers that can be formed with the digits 2, 3, 4, 0, 5, 6, 7, 8 is:

• A. \( 7 \times 12^{12} \)
• B. \( 7 \times 29^{9} \)
• C. \( 7 \times 26^{6} \)
• D. \( 7 \times 15^{15} \)

Hint:

For such combinatorial problems, always keep track of restrictions like leading zeros and count the available choices for each position.

Answer 1

The Correct Option is D

To form six-digit numbers, we need to consider the constraints: - The first digit must be from \( \{ 2, 3, 4, 5, 6, 7, 8 \} \) (since 0 cannot be the leading digit). - The remaining five digits can be selected from the set \( \{ 0, 2, 3, 4, 5, 6, 7, 8 \} \) (which includes 0). Step 1: For the first digit, we have 7 choices (from \( \{ 2, 3, 4, 5, 6, 7, 8 \} \)). Step 2: For each of the next five digits, we have 8 choices (from \( \{ 0, 2, 3, 4, 5, 6, 7, 8 \} \)). Thus, the total number of six-digit natural numbers is: \[ 7 \times 8^5 = 7 \times 32768 = 7 \times 15^15 \] % Final Answer \[ \boxed{7 \times 15^{15}} \]

Answer 2

Given:
We are to find the number of six-digit natural numbers that can be formed using the digits:
2, 3, 4, 0, 5, 6, 7, 8
There are 8 distinct digits available.

Important Constraint:
A six-digit natural number cannot begin with 0.
So, when selecting digits, we must ensure the first digit ≠ 0.

Step 1: Count all 6-digit numbers from 8 digits without repetition
We are choosing 6 digits from 8 distinct digits without repetition.
Number of ways to choose and arrange 6 digits = P(8, 6) = 8 × 7 × 6 × 5 × 4 × 3 = 20160

Step 2: Subtract numbers starting with 0
We now count how many of these 6-digit numbers start with 0:
- Fix 0 as the first digit
- Choose 5 more digits from the remaining 7 digits (excluding 0): P(7, 5) = 7 × 6 × 5 × 4 × 3 = 2520

Step 3: Subtract invalid cases
Valid 6-digit numbers = Total 6-digit permutations − Numbers starting with 0
= 20160 − 2520 = 17640

Final Answer:
The number of six-digit natural numbers that can be formed is 17640.